Hydrogen atom in its ground state is excited by means of a monochromatic radiation of wavelength `970.6 Å`. Different wavelengths are possible in the spectrum. After absorbing the energy of radiation, hydrogen atom goes to the excited state. After `10^(-8)` s, the hydrogen atom will come to the ground state by emitting the absorbed energy.
Energy absorbed by hydrogen atom is

Options

A. 1.82 eV
B. 2.18 eV
C. 8.12 eV
D. 12.8 eV

Answer

To find the energy absorbed by a hydrogen atom when it is excited by monochromatic radiation of wavelength \(970.6 \, \text{Å}\), we can use the formula that relates energy to wavelength:

\[ E = \frac{hc}{\lambda} \]

where:
- \(E\) is the energy,
- \(h\) is Planck's constant (\(6.67 \times 10^{-34} \, \text{J s}\)),
- \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)),
- \(\lambda\) is the wavelength in meters.

### Step 1: Convert the wavelength from Ångströms to meters
Given the wavelength is \(970.6 \, \text{Å}\):
\[ \lambda = 970.6 \, \text{Å} = 970.6 \times 10^{-10} \, \text{m} \]

### Step 2: Substitute the values into the energy formula
Now, we can substitute the values of \(h\), \(c\), and \(\lambda\) into the energy formula:
\[ E = \frac{(6.67 \times 10^{-34} \, \text{J s}) \times (3 \times 10^8 \, \text{m/s})}{970.6 \times 10^{-10} \, \text{m}} \]

### Step 3: Calculate the energy in Joules
Calculating the numerator:
\[ 6.67 \times 10^{-34} \times 3 \times 10^8 = 2.001 \times 10^{-25} \, \text{J m} \]
Now, divide by the wavelength:
\[ E = \frac{2.001 \times 10^{-25}}{970.6 \times 10^{-10}} \approx 2.063 \times 10^{-16} \, \text{J} \]

### Step 4: Convert the energy from Joules to electron volts
To convert Joules to electron volts, we use the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\):
\[ E_{\text{eV}} = \frac{2.063 \times 10^{-16}}{1.6 \times 10^{-19}} \approx 12.89 \, \text{eV} \]

### Final Answer
The energy absorbed by the hydrogen atom is approximately \(12.89 \, \text{eV}\).

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